Solving the ODE $y+xy'=x^4 (y')^2$ I am trying to get to the solution which is $$y=c^2 +\frac{c}{x}$$
How would I go about solving this? 
 A: $$y+x\frac{dy}{dx}=x^4\left(\frac{dy}{dx}\right)^2$$
Change of variable : $x=\frac{1}{X}$
$\frac{dy}{dx}=\frac{dy}{dX}\frac{dX}{dx}=-X^2\frac{dy}{dX}$
$$y-xX^2\frac{dy}{dX}=x^4\left(X^2\frac{dy}{dX}\right)^2$$
$$y-X\frac{dy}{dX}=\left(\frac{dy}{dX}\right)^2$$
On can see immediately that $y=aX+b$ is convenient :
$(aX+b)-aX=a^2$ imply $b=a^2$ hense $y=aX+a^2$
$$y=\frac{a}{x}+a^2$$ 
This is the expected result.
If we don't see the short-cut above, 
$$y+\frac{X^2}{4}=\left(\frac{dy}{dX}+\frac{X}{2}\right)^2$$
$$y+\frac{X^2}{4}=\left(\frac{d}{dX}\left(y+\frac{X^2}{4}\right)\right)^2$$
Let $Y=y+\frac{X^2}{4}$
$$Y=\left(\frac{dY}{dX}\right)^2$$
leading to $Y=\left(\frac{X}{2}+c\right)^2=y+\frac{X^2}{4}$
$$y=cX+c^2=\frac{c}{x}+c^2$$
A: Let $z(x) = xy(x)$, then the equation rewrites
$$z'= x^2(z'- z/x)^2.$$
We differentiate it once again:
$$z''= 2x(z'- z/x) ((z'- z/x) +x (z'' -z'/x+z/x^2)),$$
which leads to 
$$z''(1-2x^2(z'-z/x))=0.$$
If $z''=0$, then $z=ax+b$, now it is easy to put these constants in the initial equation and conclude.
If $1-2x^2(z'-z/x) = 0$, then you obtain $y = c-1/x^2$ which can never satisfy the initial equation.
A: we will show that $$y + xy' = x^4(y')^2 \tag 1  $$
has no non trivial solution in any interval containing $0$
as TZakressvky says in his post. here is another way to see that. suppose there is a solution $y$ defined in an open interval containing $0.$ substituting $x = 0,$  in $(1)$
gives $y(0) = 0.$  solve the quadratic equation $(1)$ for $y'$ gives 
$$y' = \frac1{2x^3} + \frac{\sqrt{x^2 + 4x^4}}{2x^4} , 
y' = \frac1{2x^3} -\frac{\sqrt{x^2 + 4x^4}}{2x^4}.$$ the first one cannot hold near $x = 0$ because it has a non integrable singularity of the form $\frac2{x^3}.$
we will show that second one with the minus sign cannot hold either. the right hand side $\frac1{2x^3} -\frac{\sqrt{x^2 + 4x^4}}{2x^4}$ has the opposite sign of $x.$ therefore $y$ cannot take any value other than zero in the interval containing $0.$
